Cho hai số phức z 1 , z 2 thoả mãn | z 1 | = | z 2 | = | z 1 + z 2 | = 2 . Khi đó ∣ z 1 − z 2 ∣ bằng

Hướng dẫn giải:

Đặt \({z_1} = {a_1} + {b_1}i,\,\,{z_2} = {a_2} + {b_2}i\,\,\left( {{a_1},{a_2},{b_1},{b_2} \in \mathbb{R}} \right)\). Ta có:

\(\left| {{z_1}} \right| = \left| {{z_2}} \right| = 2 \Leftrightarrow a_1^2 + b_1^2 = a_2^2 + b_2^2 = 4\)

\(\left| {{z_1} + {z_2}} \right| = 4 \Rightarrow {\left( {{a_1} + {a_2}} \right)^2} + {\left( {{b_1} + {b_2}} \right)^2} = 4\)

\( \Leftrightarrow a_1^2 + a_2^2 + 2{a_1}{a_2} + b_1^2 + b_2^2 + 2{b_1}{b_2} = 4\)

\( \Rightarrow {a_1}{a_2} + {b_1}{b_2} =  - 2\)

Do đó \({z_1} - {z_2} = \left( {{a_1} - {a_2}} \right) + \left( {{b_1} - {b_2}} \right)i\)

\(\left| {{z_1} - {z_2}} \right| = \sqrt {{{\left( {{a_1} - {a_2}} \right)}^2} + {{\left( {{b_1} - {b_2}} \right)}^2}} \)

\(\sqrt {\left( {a_1^2 + b_1^2} \right) + \left( {a_2^2 + b_2^2} \right) - 2\left( {{a_1}{a_2} + {b_1}{b_2}} \right)}  = \sqrt {4 + 4 - 2.( - 2)}  = 2\sqrt 3 \).

 Chọn A